fundamental theorem of demography, proof of

$\bullet$ First we will prove that there exist $m,M>0$ such that

$$m\le \frac{\parallel x_{k+1}\parallel }{\parallel x_k\parallel }\le M$$(1)

for all $k$,with $m$ and $M$ of the sequence. In to show this weuse the primitivity of the matrices $A_k$ and $A_{\mathrm{\infty }}$.Primitivity of $A_{\mathrm{\infty }}$ implies that there exists$l\in \mathbb{N}$ such that

$$A_{\mathrm{\infty }}^l\gg 0$$

By continuity, this implies that there exists $k_0$ such that, for all$k\ge k_0$, we have

$$A_{k+l}{A}_{k+l-1}\mathrm{\cdots }{A}_k\gg 0$$

Let us then write $x_{k+l+1}$ as a function of $x_k$:

$$x_{k+l+1}=A_{k+l}\mathrm{\cdots }{A}_k{x}_k$$

We thus have

$$\parallel x_{k+l+1}\parallel \le C^{l+1}\parallel x_k\parallel$$(2)

But since the matrices $A_{k+l}$,…,$A_k$ are strictly positivefor $k\ge k_0$, there exists a $\epsilon >0$ such that each of these matrices is superior or equal to$\epsilon$. From this we deduce that

$$\parallel x_{k+l+1}\parallel \ge \epsilon \parallel x_k\parallel$$

for all $k\ge k_0$.Applying (2), we then have that

$$C^l\parallel x_{k+1}\parallel \ge \epsilon \parallel x_k\parallel$$

which yields

$$\parallel x_{k+1}\parallel \ge \frac{\epsilon }{C^l}\parallel x_k\parallel$$

for all $k\ge 0$,and so we indeed have (1).

$\bullet$ Let us denote by $e_k$ the (normalised) Perron eigenvector of$A_k$. Thus

$$A_k{e}_k={\lambda }_k{e}_k\mathit{\hspace{1em}}\parallel e_k\parallel =1$$

Let us denote by ${\pi }_k$ the projection on the supplementary space of$\{e_k\}$ invariant by $A_k$. Choosing a proper norm, we can find$\epsilon >0$ such that

$$\left|A_k{\pi }_k\right|\le ({\lambda }_k-\epsilon )$$

for all $k$.$\bullet$ We shall now prove that

In order to do this, we compute the inner product of the sequence$x_{k+1}=A_k{x}_k$ with the $e_k$’s:

$=$

$=$

Therefore we have

$\bullet$ Now assume

We will verify that $u_k\to 0$ when $k\to \mathrm{\infty }$. We have

$u_{k+1}$$=$

and so

We deduce that there exists $k_1\ge k_0$ such that, for all $k\ge k_1$

$$|u_{k+1}|\le {\delta }_k+({\lambda }_{\mathrm{\infty }}-\frac{\epsilon }{2})|u_k|$$

where we have noted

$${\delta }_k=({\pi }_{k+1}-{\pi }_k)C^{\prime }$$

We have ${\delta }_k\to 0$ when $t\to \mathrm{\infty }$, we thus finally deduce that

$$|u_k|\to 0\text{ when }k\to \mathrm{\infty }$$

Remark that this also implies that

$$z_k=\frac{{\pi }_k{x}_k}{\parallel x_k\parallel }\to 0\text{ when }k\to \mathrm{\infty }$$

$\bullet$ We have $z_k\to 0$ when $k\to \mathrm{\infty }$, and $x_k/\parallel x_k\parallel$can be written

$$\frac{x_k}{\parallel x_k\parallel }={\alpha }_k{e}_k+z_k$$

Therefore, we have ${\alpha }_k{e}_k\to 1$ when $k\to \mathrm{\infty }$, which impliesthat ${\alpha }_k$ tends to 1, since we have chosen $e_k$ to benormalised (i.e.,$\parallel e_k\parallel =1$).

We then can conclude that

$$\frac{x_k}{\parallel x_k\parallel }\to e_{\mathrm{\infty }}\text{ when }k\to \mathrm{\infty }$$

and the proof is done.